On Friday, November 17, 2023 at 6:07:43 AM UTC-4, WM wrote:
> On 16.11.2023 21:18, William wrote:
> >
> >
> >> I specify A_1, A_2, A_3, ...
> >
> > Correct. Each is an |Nx|N matrix with the upper left corner composed of elements of |N and it is this part that is used to specify the elements of B . You cannot use this sequence to specify the elements of a putative larger matrix of which B is a proper submatix.
> I do not specify the larger matrix. Only B is specified

Correct. Your putativie matrix A that is larger than B is not specified.
So we cannot conclude anything about A (e.g. that A exists, or that B is proper submatrix of A).
The statement "no fractions leaves the matrix " is true for each A(n). but "no fraction sits at an indexed position" Is false for each A(n). No "fraction" leaves B, but no "fraction" enters B. Your game of billiards tells you absolutely nothing.

On 17.11.2023 14:49, William wrote:
> On Friday, November 17, 2023 at 6:07:43 AM UTC-4, WM wrote:
>> On 16.11.2023 21:18, William wrote:
>>>
>>>
>>>> I specify A_1, A_2, A_3, ...
>>>
>>> Correct. Each is an |Nx|N matrix with the upper left corner composed of elements of |N and it is this part that is used to specify the elements of B . You cannot use this sequence to specify the elements of a putative larger matrix of which B is a proper submatix.
>> I do not specify the larger matrix. Only B is specified
>
> Correct. Your putativie matrix A that is larger than B is not specified.

Small wonder. All positions except a small part of indexed positions -
by far not all indexed positions - are dark and therefore cannot be
specified.

> So we cannot conclude anything about A (e.g. that A exists, or that B is proper submatrix of A).

All of Cantor's B is produced because the matrices mimic Cantor's
procedure precisely. His formula k = (m + n - 1)(m + n - 2)/2 + m is
executed. Why should I deviate from his prescription?

> The statement "no fractions leaves the matrix " is true for each A(n). but "no fraction sits at an indexed position" Is false for each A(n).

Which fraction sits at an indexed position??? By the rules of the game
every fraction is replaced with the index.

> No "fraction" leaves B, but no "fraction" enters B.

B is produced by the indices which settle there. It grows step by step.
In fact only the B_n are produced. B is only produced if we trust in
Cantors conclusion from every to all.

> Your game of billiards tells you absolutely nothing.

Then let us talk about the difference in
All indices are applied to matrix positions
and
All indices are applied to roots of polynomials.

Regards, WM

On Friday, November 17, 2023 at 12:17:56 PM UTC-4, WM wrote:
> B is produced because the matrices

There are no matrices. There is an algorithm ( k = (m + n - 1)(m + n - 2)/2 + m) which give an element of |N (k) for each element of |Nx|N (m,n). Your billiards game does not give an algorithm to produce your putative matrix A.

On 17.11.2023 17:49, William wrote:
> On Friday, November 17, 2023 at 12:17:56 PM UTC-4, WM wrote:
>> B is produced because the matrices
>
> There are no matrices. There is an algorithm ( k = (m + n - 1)(m + n - 2)/2 + m) which give an element of |N (k) for each element of |Nx|N (m,n).

That is a matrix.

>Your billiards game does not give an algorithm

Of course it does! For every natural number n the indexed position is
well-defined by the rules of my game.

It is better defined, less tedious, than the indexing of the algebraic
numbers.

Regards, WM

On Friday, November 17, 2023 at 5:50:49 PM UTC-4, WM wrote:
> On 17.11.2023 17:49, William wrote:
> > On Friday, November 17, 2023 at 12:17:56 PM UTC-4, WM wrote:
> >> B is produced because the matrices
> >
> > There are no matrices. There is an algorithm ( k = (m + n - 1)(m + n - 2)/2 + m) which give an element of |N (k) for each element of |Nx|N (m,n)..
> That is a matrix.

Ok. But note it is a single algorithm/matrix, not a sequence of algorithms/matrices B is defined by (is) a single algorithm

> >Your billiards game does not give an algorithm
> Of course it does! For every natural number n the indexed position is
> well-defined by the rules of my game.

You can use the A(n) to define an algorithm which maps |N to |Nx|N (n maps to the position of n in A(n) ). This algorithm is the inverse of
k = (m + n - 1)(m + n - 2)/2 + m so you get B. Your billiards game does not define an algorithm for your putative larger matrix A.

On 18.11.2023 00:13, William wrote:
> On Friday, November 17, 2023 at 5:50:49 PM UTC-4, WM wrote:
>> On 17.11.2023 17:49, William wrote:
>>> On Friday, November 17, 2023 at 12:17:56 PM UTC-4, WM wrote:
>>>> B is produced because the matrices
>>>
>>> There are no matrices. There is an algorithm ( k = (m + n - 1)(m + n - 2)/2 + m) which give an element of |N (k) for each element of |Nx|N (m,n).
>> That is a matrix.
>
> Ok. But note it is a single algorithm/matrix, not a sequence of algorithms/matrices B is defined by (is) a single algorithm

So is A, because I use this single algorithm:
k = (m + n - 1)(m + n - 2)/2 + m.
In your eyes it entitles Cantor to produce the complete matrix B.
Therefore it entitles me to produce the complete matrix B too.
There is no difference.
But I prove that not all places get enumerated because the fractions
remain at positions inside and not enumerated.

>>> Your billiards game does not give an algorithm
>> Of course it does! For every natural number n the indexed position is
>> well-defined by the rules of my game.
>
> You can use the A(n) to define an algorithm which maps |N to |Nx|N (n maps to the position of n in A(n) ). This algorithm is the inverse of
> k = (m + n - 1)(m + n - 2)/2 + m so you get B. Your billiards game does not define an algorithm for your putative larger matrix A.

So it is. Only B is defined. But we know that the matrix which Cantor
claims to enumerate completely is not completely enumerated. He forgot
that the positions of the used indices must be indexed after the indices
have gone elsewhere.

Things are very simple: By producing the sequence 1/1, 1/2, 2/1, 1/3,
2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4,
4/3, 5/2, 6/1, ... Cantor claims completeness. He infers "all" from
"every".

By looking at the geometrical aspect, i.e., the matrix

XOOO...
XOOO...
XOOO...
XOOO...
....

we see with naked eye that his conclusion is wrong.

Regards, WM

On Friday, November 17, 2023 at 5:49:39 PM UTC+1, William wrote:

> There is an algorithm (k = (m + n - 1)(m + n - 2)/2 + m) which ...

This is not an algorithm but a formula. Aktually, the formula/expression is just "(m + n - 1)(m + n - 2)/2 + m".

On Saturday, November 18, 2023 at 11:03:12 AM UTC+1, WM wrote:

> Regards, WM

Außer William scheint niemand mehr auf Deinen psychoseiduzierten Quatsch reinzufallen, Mückenheim.

[Apart from William, no one seems to be falling for your psychotic nonsense anymore, Mückenheim.]

Please see a psychiatrist, soon.

On Friday, November 17, 2023 at 5:49:39 PM UTC+1, William wrote:

> There is an algorithm (k = (m + n - 1)(m + n - 2)/2 + m) which ...

This is not an algorithm but a formula. Aktually, the relevant expression is just "(m + n - 1)(m + n - 2)/2 + m".

On 18.11.2023 17:53, Fritz Feldhase wrote:
> On Friday, November 17, 2023 at 5:49:39 PM UTC+1, William wrote:
>
>> There is an algorithm (k = (m + n - 1)(m + n - 2)/2 + m) which ...
>
> This is not an algorithm but a formula.

In mathematics and computer science, an algorithm is a finite sequence
of rigorous instructions.
My Game of Billiards supplies it:
Push the natnumbers of the first column (without queue) into the field
of fractions and store the hit fraction always there where the natnumber
has come from. Try to push the natnumbers such that all matrix positions
are occupied by them. That is best done by creating a pattern like

B=
1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
.... ,

According to this simple rule it is impossible, in eternity, to remove a
fraction from the matrix or to attach a natnumber to a fraction.

Few instructions. Determined result.

> Aktually, the relevant expression is just "(m + n - 1)(m + n - 2)/2 + m".

One instruction is a finite sequence of instructions too.

Only B is defined. But we know that the matrix which Cantor
claims to enumerate completely
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
....
is not completely enumerated. He forgot
that the positions of the used indices must be indexed after the indices
have gone elsewhere.

Things are very simple: By producing the sequence 1/1, 1/2, 2/1, 1/3,
2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4,
4/3, 5/2, 6/1, ... Cantor claims completeness. He infers "all" from
"every".

By looking at the geometrical aspect, i.e., the matrix

XOOO...
XOOO...
XOOO...
XOOO...
....

we see with naked eye that his conclusion is wrong.

Regards, WM

On 11/18/2023 8:51 AM, Fritz Feldhase wrote:
> On Saturday, November 18, 2023 at 11:03:12 AM UTC+1, WM wrote:
>
>> Regards, WM
>
> Außer William scheint niemand mehr auf Deinen psychoseiduzierten Quatsch reinzufallen, Mückenheim.
>
> [Apart from William, no one seems to be falling for your psychotic nonsense anymore, Mückenheim.]
>
> Please see a psychiatrist, soon.

The psychiatrist better watch out! It might go crazy itself wrt
interacting with WM.

On Saturday, November 18, 2023 at 6:03:12 AM UTC-4, WM wrote:

> > You can use the A(n) to define an algorithm which maps |N to |Nx|N (n maps to the position of n in A(n) ). This algorithm is the inverse of
> > k = (m + n - 1)(m + n - 2)/2 + m so you get B. Your billiards game does not define an algorithm for your putative larger matrix A.
> So it is. Only B is defined.
Correct. You do not define the putative larger matrix A.

On 19.11.2023 00:04, William wrote:
> On Saturday, November 18, 2023 at 6:03:12 AM UTC-4, WM wrote:
>
>>> You can use the A(n) to define an algorithm which maps |N to |Nx|N (n maps to the position of n in A(n) ). This algorithm is the inverse of
>>> k = (m + n - 1)(m + n - 2)/2 + m so you get B. Your billiards game does not define an algorithm for your putative larger matrix A.
>> So it is. Only B is defined.
> Correct. You do not define the putative larger matrix A.

Correct. I repeat Cantor and define his matrix B. But I show that this
is not the whole matrix to be indexed. The larger parts cannot be
defined because they are dark. But they contain all fractions. In
mathematics we use indirect proofs, sometimes.

Regards, WM

On Sunday, November 19, 2023 at 6:10:30 AM UTC-4, WM wrote:
> On 19.11.2023 00:04, William wrote:
> > On Saturday, November 18, 2023 at 6:03:12 AM UTC-4, WM wrote:
> >
> >>> You can use the A(n) to define an algorithm which maps |N to |Nx|N (n maps to the position of n in A(n) ). This algorithm is the inverse of
> >>> k = (m + n - 1)(m + n - 2)/2 + m so you get B. Your billiards game does not define an algorithm for your putative larger matrix A.
> >> So it is. Only B is defined.
> > Correct. You do not define the putative larger matrix A.
> Correct.

So your billiard game only defines the A(n) which tell you nothing.

On 19.11.2023 12:55, William wrote:
> On Sunday, November 19, 2023 at 6:10:30 AM UTC-4, WM wrote:
>> On 19.11.2023 00:04, William wrote:

>>> Correct. You do not define the putative larger matrix A.
>> Correct.
>
> So your billiard game only defines the A(n) which tell you nothing.
>
My game of billiards is played on the field A(0) which has the same size
as every A(n) and the final state A, if the latter exists. B was
believed to be of the same size, but that belief is refuted because the
fractions are in the field but not in B.

Regards, WM

On Sunday, November 19, 2023 at 11:49:29 AM UTC-4, WM wrote:
> On 19.11.2023 12:55, William wrote:

> > So your billiard game only defines the A(n) which tell you nothing.
> >
> My game of billiards is played on the field A(0) which has the same size
> as every A(n)

Correct. This tells you nothing.

On 19.11.2023 17:17, William wrote:
> On Sunday, November 19, 2023 at 11:49:29 AM UTC-4, WM wrote:
>> On 19.11.2023 12:55, William wrote:
>
>>> So your billiard game only defines the A(n) which tell you nothing.
>>>
>> My game of billiards is played on the field A(0) which has the same size
>> as every A(n)
>
> Correct. This tells you nothing.
>
Not quite. It tells me that B is established as far as Cantor
establishes it - namely by utilizing quite the same algorithm
k = (m + n - 1)(m + n - 2)/2 + m). And it tells me that all matrices
A(i) are larger than B or whatever B(i) is finally produced by Cantor.

If it does not tell you the same, then you should try to understand better.

Reards, WM

On Sunday, November 19, 2023 at 1:09:58 PM UTC-4, WM wrote:
> it tells me that all matrices A(i) are larger than B

Nope. It tells you that all the A(i) are the same "size". It does not tell you that A(0) is "larger" than B.

On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:

> Cantor [...] define[s] his matrix B.

Right. Actually, he defined the function f: IN x IN --> IN, (n, m) |-> (m + n - 1)(m + n - 2)/2 + m), which can be interpreted as a matrix of size "IN x IN".

This function is a bijection from IN x IN onto IN.

On Sunday, November 19, 2023 at 7:10:04 PM UTC+1, Fritz Feldhase wrote:
> On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:
> >
> > Cantor [...] define[s] his matrix B.
> >
> Right. Actually, he defined the function f: IN x IN --> IN, (n, m) |-> (m + n - 1)(m + n - 2)/2 + m, which can be interpreted as a matrix of size "IN x IN".
>
> This function is a bijection from IN x IN onto IN.

Um nun die unendliche Folge zu erhalten, von der Cantor im Folgenden spricht, braucht man nur die Umkehrfunktion f^-1 von f zu betrachten. Es gilt dann:

"every [fraction] p/q comes at an absolutely fixed position of a simple infinite sequence"

und:

"The infinite sequence thus defined has the peculiar property to contain the positive [fractions] completely, and each of them only once at a determined place."

This infinite sequence starts with the following elements (in the given order):

1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

On Sunday, November 19, 2023 at 4:49:29 PM UTC+1, WM wrote:

> as every A(n) and the final state A

There is no "final state", you silly monkey.

All you have is the _infinite_ sequence (A(i))_(i e iN) of IN x IN matrices, none of which is identical with the IN x IN matrix B.

"This tells you nothing." (William)

On 11/19/2023 2:10 AM, WM wrote:
> On 19.11.2023 00:04, William wrote:
>> On Saturday, November 18, 2023 at 6:03:12 AM UTC-4, WM wrote:
>>
>>>> You can use the A(n) to define an algorithm which maps |N to |Nx|N
>>>> (n maps to the position of n in A(n) ). This algorithm is the
>>>> inverse of
>>>> k = (m + n - 1)(m + n - 2)/2 + m so you get B. Your billiards game
>>>> does not define an algorithm for your putative larger matrix A.
>>> So it is. Only B is defined.
>> Correct.   You do not define the putative larger matrix A.
>
> Correct. I repeat Cantor and define his matrix B. But I show that this
> is not the whole matrix to be indexed. The larger parts cannot be
> defined because they are dark. But they contain all fractions. In
> mathematics we use indirect proofs, sometimes.

Cantor pairing maps all of the naturals to pairs, and back again, full
circle. Your brain is in the dark. I am thinking of a number... Take a
guess, is not dark at all. lol.

On 11/19/2023 10:32 AM, Fritz Feldhase wrote:
> On Sunday, November 19, 2023 at 4:49:29 PM UTC+1, WM wrote:
>
>> as every A(n) and the final state A
>
> There is no "final state", you silly monkey.

The moron cannot understand that once he says final state, it goes into
the finite. A snapshot of an infinite process. Well, there are infinite
snapshots... WM is hyper ultra finite.

>
> All you have is the _infinite_ sequence (A(i))_(i e iN) of IN x IN matrices, none of which is identical with the IN x IN matrix B.
>
> "This tells you nothing." (William)

On 11/19/2023 10:24 AM, Fritz Feldhase wrote:
> On Sunday, November 19, 2023 at 7:10:04 PM UTC+1, Fritz Feldhase wrote:
>> On Sunday, November 19, 2023 at 11:10:30 AM UTC+1, WM wrote:
>>>
>>> Cantor [...] define[s] his matrix B.
>>>
>> Right. Actually, he defined the function f: IN x IN --> IN, (n, m) |-> (m + n - 1)(m + n - 2)/2 + m, which can be interpreted as a matrix of size "IN x IN".
>>
>> This function is a bijection from IN x IN onto IN.
>
> Um nun die unendliche Folge zu erhalten, von der Cantor im Folgenden spricht, braucht man nur die Umkehrfunktion f^-1 von f zu betrachten. Es gilt dann:
>
> "every [fraction] p/q comes at an absolutely fixed position of a simple infinite sequence"
>
> und:
>
> "The infinite sequence thus defined has the peculiar property to contain the positive [fractions] completely, and each of them only once at a determined place."
>
> This infinite sequence starts with the following elements (in the given order):
>
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

Fwiw, its fun to use these mapped Cantor pairs as notes for some music.

https://youtu.be/XkwgJt5bxKI

When you get really bored, give it a go. ;^)

On Sunday, November 19, 2023 at 8:58:47 PM UTC+1, Chris M. Thomasson wrote:

> WM is hyper ultra finite.

You mean his brain? Hyper ultra finite and tiny.